How is pole-polar duality related to line-point duality.

Question

The above image is my depiction of what I am calling line-point duality. We can either think of the "point" as the equivalence class of arrows parallel to the large silver arrow ($\mathfrak{L}$ drawn as a vector), or as the intersection of that arrow with the projective plane, drawn one unit above the $X\times{Y}$ plane. The corresponding "line" can be understood as the plane normal to the arrow representation of $\mathfrak{L},$ or as the intersection of that plane with the projective plane. In the following list of expressions, the first line shows the relation ship of the rational form notation to my notation, which appears in the illustration. The second line is the Cartesian form of the line in $\mathbb{R}^2.$ The third line is the equation of the plane in $\mathbb{R}^3.$ The symbol $\hat{\mathfrak{e}}_{o}$ is the standard basis vector normal to the $X\times{Y}$ plane.

\begin{align*} \mathfrak{L}= &\left\{ L_{x}:L_{y}:L_{o}\right\}\cong \tilde{\mathfrak{L}}+L_{o}\hat{\mathfrak{e}}^{o}\\ 0=&L_{x}x+L_{y}y+L_{o}\\ 0=&L_{x}P^{x}+L_{y}P^{y}+L_{o}P^{o} \end{align*}

The following image was cribbed from one of Norman Wildberger's excellent YouTube videos. If I can find the URL, I will add it. It shows what is sometimes called pole-polar duality, and what Wildberger is calling Apollonius' duality, at this point in his lecture.

It seems there must be some connection between these two kinds of duality. Note that, if we assume the origin is at the center of the circle in Wildberger's drawing, then the line (polar) and point (pole) lie on the "same side", whereas, in my drawing, the point and line are on "opposite sides" of the origin. If the pole is moved toward the circle, the polar moves toward the pole. In my illustration, as the point moves toward the origin, the line moves off to infinity.

So, is there a connection between these two kinds of duality between lines and points?

Answer 1

$ \renewcommand\propto\approx $2D Plane-based Geometric Algebra (PGA) shows the relationship between these nicely. Sadly I don't think this is very accessible. I learned from Charles Gunn's 2011 PhD thesis, though maybe some videos like these would be a better start. Even then, you should really start by learning geometric algebra first; I would recommend either Doran and Lasenby's Geometric Algebra for Physicists (2003) or Lounesto's Clifford Algebras and Spinors (2001).

2D PGA is a formalism where points, lines, and free vectors (and the plane itself) are algebraic objects. It is a projective model, so multiplying by a non-zero scalar doesn't change the geometric object represented, and there is a line at infinity $\infty$. Free vectors, representing direction purely, are points on $\infty$. I will use $\propto$ to denote that two algebraic objects represent the same geometric object, i.e. are equal up to a non-zero scalar multiple.

If $C, P$ are mutually oriented points so that $P + C$ is a point on the line segment between them, then the line $p$ line-point dual to $P$ relative to the center $C$ is $$ p = PI + (P\vee C)C^{-1}. $$ $I$ is the pseudoscalar (which could maybe be thought of as representing the projection point of the projective space); $PI \propto \infty$ for any point $P$. The product $\vee$ is the join product so that $P\vee C$ is the line passing through those points. $C^{-1} \propto C$, and the inverse is necessary so that the magnitude of $C$ does not effect the value of $p$.

If $C$ is the center of a unit circle, then the line $p$ pole-polar dual to $P$ is $$ p = PI - (P\vee C)C^{-1}. $$ That's it! If the circle instead has radius $R$, we can scale the whole space to get the correct formula. It turns out that $I \mapsto RI$ and every $\vee$ picks up a factor of $1/R$ so that $$ p = PIR - \frac{P\vee C}RC^{-1} \propto R^2PI - (P\vee C)C^{-1}. $$

Some calculation will show that if $v_\perp$ is the line through $C$ perpendicular to $P\vee C$ and the distance beween $P$ and $C$ is $d$ that $$ p = PI \pm (P\vee C)C^{-1} \propto \infty \mp dv_\perp \propto v_\perp \mp \frac1d\infty, $$ showing that $p$ is $v_\perp$ shifted a distance $\mp1/d$ towards $P$ (away if negative). The scaled formula gives $$ p \propto v_\perp + \frac{R^2}d\infty $$ as it should.

The same exact formula works when $P$ is replace with a line $l$ to get its dual point $L$: $$ L = lI \pm (l\vee C)C^{-1}. $$ $lI$ is the the free vector perpendicular to $l$, and $l\vee C$ is the entire plane (represented by a non-zero scalar).